Angular Momentum: mud hitting door

[PirateFan308]

Homework Statement

A solid wood door 1.00m wide and 2.00m high is hinged along one side and has a total mass of 42.0kg. Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.500kg, traveling perpendicular to the door at 14.0m/s just before impact.
Find the angular speed of the door.

Homework Equations

L_{initial}=L_{final}

L=Iw=rp

The Attempt at a Solution

L_{initial}=L_{door}+L_{mud}=0+mvr=(0.5)(14)(0.5)=3.5

L_{final}=I_{total}+w_{f}

I_{total}=I_{door}+I_{mud}

I'm not sure how to calculate I_{door}, which gets me stuck here.
I had read online that it is:

I_{door}=\frac{1}{3}(42)(0.5)^2=3.5

Mud I know:

I_{mud}=Iw=(0.5)(0.5)^2=1.125

Assuming I calculated the door correctly,

L_{initial}=L_{final}

3.5=(3.5+0.125)w_{f}

w_{f}=0.966

But this is wrong, and I'm not sure where I messed up. Thanks in advance!

 

[PirateFan308]

I figured out the correct answer, but in order to get it, I needed to say that

I_{door}=\frac{1}{3}(42)(1)^{2}

I don't understand why I should use 1 for the radius of the door, when the force was applied to the centre of the door.